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Indefinite Integrals question

2021 · 27 Aug · Shift 2 · Q41
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Indefinite Integrals question

2021 · 27 Aug · Shift 2 · Q41

JEE MainMathematicsIndefinite IntegralsNumerical+4 / −1
If ∫2ex+3e−x4ex+7e−xdx=114(ux+vlog⁡e(4ex+7e−x))+C\int {{{2{e^x} + 3{e^{ - x}}} \over {4{e^x} + 7{e^{ - x}}}}dx = {1 \over {14}}(ux + v{{\log }_e}(4{e^x} + 7{e^{ - x}})) + C}∫4ex+7e−x2ex+3e−x​dx=141​(ux+vloge​(4ex+7e−x))+C, where C is a constant of integration, then u + v is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 7

  1. We need to evaluate
I=\int \frac{2e^x+3e^{-x}}{4e^x+7e^{-x}}\,dx

and match it with

I=114(ux+vln⁡(4ex+7e−x))+C.I=\frac{1}{14}\left(ux+v\ln(4e^x+7e^{-x})\right)+C.I=141​(ux+vln(4ex+7e−x))+C.
  1. Let
D=4ex+7e−x.D=4e^x+7e^{-x}.D=4ex+7e−x.

Then

D′=4ex−7e−x.D'=4e^x-7e^{-x}.D′=4ex−7e−x.

We now try to express the numerator as a linear combination of DDD and D′D'D′:

2ex+3e−x=A(4ex+7e−x)+B(4ex−7e−x).2e^x+3e^{-x}=A(4e^x+7e^{-x})+B(4e^x-7e^{-x}).2ex+3e−x=A(4ex+7e−x)+B(4ex−7e−x).

Comparing coefficients:

4A+4B=2,4A+4B=2,4A+4B=2, 7A−7B=3.7A-7B=3.7A−7B=3.

So,

A+B=12,A+B=\frac12,A+B=21​, A−B=37.A-B=\frac37.A−B=73​.

Adding,

2A=12+37=7+614=1314  ⟹  A=1328.2A=\frac12+\frac37=\frac{7+6}{14}=\frac{13}{14} \implies A=\frac{13}{28}.2A=21​+73​=147+6​=1413​⟹A=2813​.

Then

B=12−1328=14−1328=128.B=\frac12-\frac{13}{28}=\frac{14-13}{28}=\frac{1}{28}.B=21​−2813​=2814−13​=281​.

Thus,

2ex+3e−x=1328(4ex+7e−x)+128(4ex−7e−x).2e^x+3e^{-x}=\frac{13}{28}(4e^x+7e^{-x})+\frac{1}{28}(4e^x-7e^{-x}).2ex+3e−x=2813​(4ex+7e−x)+281​(4ex−7e−x).

Dividing by DDD,

2ex+3e−x4ex+7e−x=1328+128⋅4ex−7e−x4ex+7e−x.\frac{2e^x+3e^{-x}}{4e^x+7e^{-x}}=\frac{13}{28}+\frac{1}{28}\cdot\frac{4e^x-7e^{-x}}{4e^x+7e^{-x}}.4ex+7e−x2ex+3e−x​=2813​+281​⋅4ex+7e−x4ex−7e−x​.
  1. Therefore,
I=∫[1328+1284ex−7e−x4ex+7e−x]dx.I=\int \left[\frac{13}{28}+\frac{1}{28}\frac{4e^x-7e^{-x}}{4e^x+7e^{-x}}\right]dx.I=∫[2813​+281​4ex+7e−x4ex−7e−x​]dx.

So,

I=1328x+128∫4ex−7e−x4ex+7e−xdx.I=\frac{13}{28}x+\frac{1}{28}\int \frac{4e^x-7e^{-x}}{4e^x+7e^{-x}}dx.I=2813​x+281​∫4ex+7e−x4ex−7e−x​dx.

Since D′=4ex−7e−xD'=4e^x-7e^{-x}D′=4ex−7e−x,

∫4ex−7e−x4ex+7e−xdx=ln⁡(4ex+7e−x).\int \frac{4e^x-7e^{-x}}{4e^x+7e^{-x}}dx=\ln(4e^x+7e^{-x}).∫4ex+7e−x4ex−7e−x​dx=ln(4ex+7e−x).

Hence,

I=1328x+128ln⁡(4ex+7e−x)+C.I=\frac{13}{28}x+\frac{1}{28}\ln(4e^x+7e^{-x})+C.I=2813​x+281​ln(4ex+7e−x)+C.
  1. Match this with
I=114(ux+vln⁡(4ex+7e−x))+C.I=\frac{1}{14}(ux+v\ln(4e^x+7e^{-x}))+C.I=141​(ux+vln(4ex+7e−x))+C.

That is,

u14=1328,v14=128.\frac{u}{14}=\frac{13}{28}, \qquad \frac{v}{14}=\frac{1}{28}.14u​=2813​,14v​=281​.

So,

u=132,v=12.u=\frac{13}{2}, \qquad v=\frac{1}{2}.u=213​,v=21​.

Therefore,

u+v=132+12=7.u+v=\frac{13}{2}+\frac{1}{2}=7.u+v=213​+21​=7.
  1. Final answer:
7\boxed{7}7​
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